Two questions about trig functions of matrices So the

Magdalena Warren

Magdalena Warren

Answered question

2022-04-09

Two questions about trig functions of matrices
So the first part of this question asked me to define cosA and sinA where A is a nxn real matrix. I used the taylor series expansions of sin and cos to get
sinA=n=0(1)n(2n+1)!A2n+1
cosA=n=0(1)n(2n)!A2n
For part a. I need to show that
ddtsin(At)=Acos(At)
ddtcos(At)=Asin(At)
I was able to do this for sin with the following steps:
ddtsin(At)=ddt(n=0(1)n(2n+1)!(At)2n+1)=ddt(n=0(1)n(2n+1)!A2n+1t2n+1)=n=0(1)n(2n+1)!A2n+1(2n+1)t2n=n=0(1)n(2n)!A2n+1t2n=An=0(1)n(2n+1)!(At)2n=Acos(At)
However, for cosine I get stuck. I start off the same way:

Answer & Explanation

firenzesunzc65

firenzesunzc65

Beginner2022-04-10Added 16 answers

Step 1
Instead of focusing on a particular function, consider a generic analytic function and its derivative.
f(x)=k=0βkxk      g(x)=dfdx=k=0k βkxk1f˙=(dfdx)(dxdt)=gx˙
Now apply these functions to a square matrix
F=f(X)=k=0βkXk    G=g(X)=k=0k βkXk1F˙=k=0βkj=0k1Xj X˙ Xkj1  GX˙ 
The problem is that (in general) X˙ does not commute with X.
Step 2
However, given a constant matrix A, the matrix X=At does commutes with its time derivative since
X˙=A    XX˙=X˙X=A2t
Substituting this X into the expression for F˙ allows terms to be collected exactly they way they were in the scalar case, yielding F˙=GA=AG.
Step 3
In this particular problem, f(x)=sin(x) and g(x)=cos(x)
For part b, if you are given a functional identity
f1(x)+f2(x)=f3(x)
it doesn't matter if you evaluate the LHS or the RHS, because they are identical.
You merely need to recognize the RHS as a function, which can be translated into matrix form
x0=1  X0=I
Or if you prefer, you can think of it as a Taylor series with a single term
f3(x)=(sin2(x)+cos2(x))=(k=00xk)=1

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